Answer to Question #346621 in Math for Busi

Question

An aeroplane flies at a ground velocity (i.e. velocity relative to the ground) of 300 km/h N 30o W, in a wind blowing at a velocityof 50 km/h N 20o E. What is the velocity (speed and direction)of the plane relative to the ground? (Use a calculator and round the speed)

to the nearest km/h, and the corresponding angle to the nearest degree.)

Answer 1

$$\vec{v_p}=300\cos(90\degree+30\degree)\vec{i}+300\sin(90\degree+30\degree)\vec{j}$$

$$\vec{v_w}=50\cos(90\degree-20\degree)\vec{i}+50\sin(90\degree-20\degree)\vec{j}$$

$$\vec{v}=\vec{v_p}+\vec{v_w}=300\cos(120\degree)\vec{i}+300\sin(120\degree)\vec{j}$$

$$+50\cos(70\degree)\vec{i}+50\sin(70\degree)\vec{j}$$

$$|\vec{v}|^2=(300\cos(120\degree)+50\cos(70\degree))^2$$

$$+(300\sin(120\degree)+50\sin(70\degree))^2$$

$$\approx111783.6283$$

$$|\vec{v}|=\sqrt{|\vec{v}|^2}\approx334\ km/h$$

$$\tan \theta=\dfrac{300\sin(120\degree)+50\sin(70\degree)}{300\cos(120\degree)+50\cos(70\degree)}$$

$$\approx-2.30846$$

$$\theta=180\degree+\tan^{-1}(-2.30846)\approx113\degree$$

$334\ km/h$, N $23\degree$W